Stack Books at the Edge of a Table

Use the following method of stacking: The top book extends half its length beyond the second book. The second book extends a quarter of its length beyond the third. The third extends one-sixth of its length beyond the fourth, and so on. Let $m$ and $L$ be the mass and length of each book, respectively. And take the edge of the table as the origin of number line as shown. When we have $n$ books, denote the center of mass of $n$ books by $c_n$ , and the distance that the top book extends beyond the edge of the table $d_n$ . When there are $(n+1)$ books, by the method of stacking, we get $\begin{aligned} c_{n+1}&=\frac{nm\left(c_n+\frac L{2n}\right)+m\left(-\frac L2\right)}{(n+1)m} \\&=\frac n{n+1}c_n \\&=\frac n{n+1}\frac{n-1}n\ldots\frac 12c_1 \\&=-\frac L{2(n+1)}<0. \end{aligned}$ Therefore, these books couldn't collapse. In this case, we have $d_{n+1}=\frac L2\left(1+\frac12+\frac13+\ldots+\frac1n\right).$ Since harmonic series $\sum_{k=1}^\infty\frac1k$ is divergent, the top book can extend any distance at all beyond the edge of the table as long as the stack is high enough, that is, the answer is $\boxed{\text{Yes}}$ .